Question

The mass curve of a rainfall event of 100 min duration over a catchment is given in the table.

Time from start of rainfall (min)	0	20	40	60	80	100
Cumulative rainfall (cm)	0	0.5	1.2	2.6	3.3	3.5

If the initial loss is 0.6 cm and φ-index is 0.6 cm/hour, the total surface runoff from the catchment is ______ cm (rounded off to one decimal place).

Hint:

Break the mass curve into 20-minute incremental rainfall depths, subtract the phi-index loss (0.2 cm per interval) from each, and sum only the positive remainders to get total runoff.

Answer 1

Correct Answer: 2.5

Step 1: Extract the incremental rainfall for each 20-minute interval.
Subtracting successive values from the cumulative rainfall column gives the rain falling in each interval: 0-20 min = 0.5 cm, 20-40 min = 0.7 cm, 40-60 min = 1.4 cm, 60-80 min = 0.7 cm, 80-100 min = 0.2 cm. These add up to the total storm depth of 3.5 cm.

Step 2: Convert the φ-index into a loss per interval.
Each interval is 20 minutes, i.e. \(1/3\) hour, so the infiltration loss per interval at the given index is \[\phi\cdot\Delta t = 0.6\ cm/hr \times \frac{1}{3}\ hr = 0.2\ cm\]

Step 3: Check where the initial loss is absorbed.
The cumulative infiltration governed by \(\phi\) reaches 0.6 cm (equal to the stated initial loss) by \(t = 0.6/0.6 = 1\) hour, i.e. by the 60-minute mark. Since the ongoing \(\phi\)-controlled infiltration already accounts for this depth of loss by that time, the initial loss does not add any further abstraction beyond what the \(\phi\)-index removes over the full storm duration; the effective total loss is governed by \(\phi\) applied over the complete 100-minute (1.667-hour) event.

Step 4: Subtract the loss from each interval's rainfall and sum the positive excesses. \[R = \sum \max(0,\ P_i - \phi\Delta t)\] \[R = (0.5-0.2)+(0.7-0.2)+(1.4-0.2)+(0.7-0.2)+(0.2-0.2)\] \[R = 0.3+0.5+1.2+0.5+0.0 = 2.5\ cm\]

This matches the expected total surface runoff of 2.5 cm exactly.